Archive for the ‘Exercises’ Category

clean-dishes montage

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domestic arts in the age of digital distribution. ┬áRIP, vlorblog. ┬áhere are some posts mentioning “clean” and “dish”: https://vlorblog.wordpress.com/?s=clean+dish (courtesy of the “search” feature from that blog).

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songwriting 201

Photo on 8-19-14 at 11.55 AM

our medium is handwriting. but
*don’t* trust even this little
bit of score; i can’t keep a beat
*even in real life*… not like a
drummer, say… with my rhythmic
body and *darn well* haven’t learned
how to put the *numbers* in.
syllable counts and beat counts just
blur up fast when i try to get a handle
on suchlike matters so i always give up
right away.

but scribbling out this display here
*was* helpful in learning to play the
damn thing out note-by-note. (i’ve been
wandering around the house singing this
for a few weeks now).

biopsy the barrier

i got nothin
to say uh…
specially about
whatever was.

you can’t?
repeat the past?
of *course* y’can!

break on thoo
t’ th’ othuh side.
brae, brea,
brea, khah-nthoo.
bray, cahhn, thoo.
bray-con-thoo.
(to,
the other side).

is this some dagger
of the *mind*?
or…, well…,
what?

leaving only the “typing it up worse
for the net” to me… the \TeX-set
homework i’m about to loosely base my
plain-text solutions on is quite a lovely
thing to look upon… as my versions will
not be. now. make it so.

6.1.3
Let a/b, a’/b’, and a”/b” be consecutive fractions
in the Farey sequence of order n.

(i.e., in
F_n = {0/1, 1/n, …, 1/2, … , (n-1)/n, 1/1}
[for the record… it’s convenient to
have a “symbol” for this sequence;
F-subscript-n is standard but the
text doesn’t include it.]
)
We will show that these fractions
satisfy the “Mediant Property”:
a’/b’ = (a+a”)/(b+b”).

Consecutive “Farey fractions”
satisfy Theorem 6.1 and so
a’b – ab’ = 1 and
a”b’ – a’b” = 1.

Equating the left-hand sides,
a’b – ab’ = a”b’ – a’b”.

Regroup; factor;
“divide through by b'(b+b”)”:
a'(b+b”) = b'(a”+a)
a’/b’ = (a+a”)/(b+b”).
(We are done.)

in one sense, “textbook” exercises.
but in the usual sense, not.
i made up these “exercises” for myself,
but they sure as heck involve a *textbook*.
just now… very much in my usual way…
i made some pencil notes on a textbook page.

there was even a “warm-up”.
my *first* marks on that page…
“page 43”, let’s say… were to
scratch over the first “of” of

In number theory, the function of given in Theorem 2.2 would be called an ordered composition of k with n parts, because it gives a list of n numbers that add up to k.
and replace it with “f“.

because theorem 2.2, as it turns out,
tells us that

Theorem 2.2. The number of functions f from a set {x_1, x_2, \ldots x_n} to the nonnegative integers such that \sum_{i=1}^n f(x_i) = k is C(n + k – 1; k).

or, in other words, that
\sharp \{ f \mid f: \{x_i\}_{i=1}^{n} \rightarrow {\Bbb N}, \sum_{i=1}^n f(x_i) = k\} ={{n+k-1}\choose k}.

and that was, if you will, “exercise 1”:
“restate theorem 2.2 wordlessly”.
this code now appears in the (generous)
margin of page 43 (with the unimaginative
label “THEOREM 2.2”).

exercise 2?
well, the
\sum_{i=1}^n f(x_i) = k
bit appears *twice* on page 43
centered on a line of its own
(in a “display”, in other words).
so next to the first such display
i wrote out
(
f TAKES
n VALUES
ADDING UP
TO k
)
… where the parens “(“,”)” indicate
big ol’ *handwritten* parens
bracketing the whole “block”
of handwritten “f takes n
values adding up to k” thing;
it fills the “display” space
(to the right of the displayed code)
rather nicely from top to bottom.

so the first “exercise” was of the form:
“translate from english into code”
and the second was of the form
“translate from code into english”.

“write it up in \TeX” will here
be considered beyond the scope of our
post
… it *is* time for breakfast…

i just couldn’t wait to observe here that such
“translating”, back-and-forth, code-to-language,
language-to-code is a *large part* of “doing math”.
throw in code-to-graph (and back) and get even more.
venn-diagrams back-and-forth to truth tables.
(hey, look. “isomorphisms”!)

anyhow. we’re *always* doing suchlike exercises.
trouble is. what-you-mean-“we”?

*feed* me!

6.
With a TI “grapher”, I’ve just found that
244^2 == 5 (mod 1009). One simply puts
Y_1 = 1009(fPart(X^2/1009))…
this computes the equivalence-class-mod-1009
of the square of each input successively…
and “scrolls down” in the TABLE window
(set to display natural number X’s).

Hence x^2 == 5 (mod 1009) *does* have a solution.
(In fact, two of them; the other is of course
765 (= 1009 – 244)).

In detail,
244^2 = 58536 = 59*1009 + 5
and
765^2 = 585225 =580*1009 + 5.
(The point here is that one may easily
*verify* such results by “paper-&-pencil”
methods.)

To find the roots *entirely* by p-&-p methods,
the best line of attack would probably be via
“primitive roots”.

trouble is, is, that that “5”
was supposed to’ve been “150”.
once more, dear friends!
(or close the wall up with our
english dead!)

handwritingNum'Th'y1

one could simply cite the lemma
and get it over with. but that’s
not how i actually did the exercise…

Photo on 2014-03-14 at 18.56

ladies and gentlemen, PSL(2,7).

Let p be an odd prime.
Let L = (1^2)(3^2)…([p-1]^2) and
let R = (-1)^{(p+1}/2}.
We will show that L = R (mod p).

Note that for 1 \le k < p
("\le" denotes is-less-than-or-equal-to)
we have k = -(p-k) (mod p).
Rearranging the factors of L gives
L=1(-1)(p-1)2(-1)(p-2)…(p-2)(-1)(2).
L=(-1)^{(p-1)/1}1*2*3*…*(p-1).
Since 1*2*…*(p-1) = (p-1)!, we can
apply Wilson's Theorem:
L\equiv (-1)^{(p-1)/2}*(-1) (mod p).
But then L = R as we were to show.

One may now derive
(2^2)(4^2)…([p-1]^2) \equiv R
by squaring "Wilson's Equation"
[(p-1)!]^2 \equiv (-1)^2 (mod p).
Note that (-1)^2 = 1 = (-1)^{p-1}:
[(p-1)!]^2 \equiv(-1)^{p-1} (mod p).

Divide on the left by L;
divide on the right by R.
The result follows.
(Division "works" because p
is a prime… we can always
"cancel" factors less than p
in this case.)

(One should *not*… if instructed
to prove *two* equivalences…
simply remark that the second
"is similar to" the first:
if our textbook authors hadn't
wanted at least *some* details,
they presumably would not have
asked that part of the question
at all.)

(n!+1, (n+1)!+1)=1. [For all n \in N.]

[Let n \in N.]
Suppose (n!+1, (n+1)!+1) = K > 1.
Then K has a prime divisor p>1.
We have p|(n!+1) and p|(n+1)!+1.
Since “p|x and p|y” implies that
“p|(y-x)” in all cases, we have
p|[ (n+1)!+1 – (n!+1) ]
p|[ (n+1)n! – n!]
p|n*n!.
But then (because p is prime) p|n!.
This is absurd since we also know p|n!+1.
No such p exists; K = 1; we are done.
[
i’ve been cranking out *lots* of number
theory problems… i’ve never taught the
subject at this level of detail and *need*
to do lots of problems (in order that
exercises like this can *become* “routine”.)
but i’ve only *typed up* a few (one
problem set [out of four that i’ve marked
so far]). anyhow, none of the class
had a proof this short-&-sweet so here
it is now for my loyal subscribers.
this is a *great* quarter so far.
getting back to work.
oops, “semester” so far.
\bye
]